Proof of Comparison Principle for the Dirichlet Problem for Second-Order Equations

Throughout, $F$, $\gamma$, $\omega$, $u$ and $v$ are as in the statement, and $T$ denotes the domain of $\omega$, the set of nonnegative real numbers.

Conventions. We use without further comment that $\le$ on $\mathbb{R}$ is transitive and compatible with addition, by the ordered field axioms and Total Order on a Set, and the weak compatibility with multiplication: if $0\le b$ and $s\le t$, then $sb\le tb$; this is immediate when $b=0$ or $s=t$, and otherwise follows from claim 10 of Elementary Order Arithmetic in an Ordered Field.

Step 1 (The doubled variational problem). Write $K=\overline{\Omega}$, which is nonempty and compact by Bounded Open Domain in Euclidean Space §closure. Equip $\mathbb{R}^{n}\times\mathbb{R}^{n}$ with the product metric $d_{\times}$ obtained from $d_{E}$ and $d_{E}$, a metric by claim 1 of The Product Metric is a Metric; then $K\times K$ is nonempty and compact in $\mathbb{R}^{n}\times\mathbb{R}^{n}$ by A Product of Compact Subsets is Compact in the Product Metric.

By Viscosity Subsolution and Supersolution up to the Boundary the function $u$ is upper semicontinuous on $\overline{\Omega}$ and $v$ is lower semicontinuous on $\overline{\Omega}$. Hence, by claim 4 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, the function $\Phi:K\times K\to\mathbb{R}$ with $\Phi(x,y)=u(x)-v(y)$ is upper semicontinuous on $K\times K$ with respect to $d_{\times}$.

Let $\psi:K\times K\to\mathbb{R}$ be given by $\psi(x,y)=d_{E}(x,y)\,d_{E}(x,y)$. The hypotheses of The Squared-Distance Penalization Limit on a Compact Set hold for the metric space $(\mathbb{R}^{n},d_{E})$, the compact set $K$ and the functions $u$ and $v$, so that corollary gives: $\psi$ is lower semicontinuous on $K\times K$ with respect to $d_{\times}$, $0\le\psi(x,y)$ for every $(x,y)\in K\times K$, and $\psi(x,y)=0$ if and only if $x=y$.

Put $\Psi=\tfrac{1}{2}\psi$, so that $\Psi(x,y)=\tfrac{1}{2}\,d_{E}(x,y)^{2}$. By claim 8 of Elementary Order Arithmetic in an Ordered Field the number $\tfrac{1}{2}$ is positive, so $\Psi$ is lower semicontinuous on $K\times K$ by claim 3 of Sums and Nonnegative Multiples of Semicontinuous Functions, satisfies $0\le\Psi(x,y)$ everywhere, and satisfies $\Psi(x,y)=0$ if and only if $\psi(x,y)=0$, that is if and only if $x=y$. Consequently the set

is nonempty.

The hypotheses of Limits of Penalised Maxima on a Compact Subset of a Metric Space therefore hold for the metric space $\bigl(\mathbb{R}^{n}\times\mathbb{R}^{n},d_{\times}\bigr)$, the compact set $K\times K$, and the functions $\Phi$ and $\Psi$. We use the notation of that lemma: for positive $\alpha\in\mathbb{R}$, $M_{\alpha}$ is the maximum value of $\Phi-\alpha\Psi$ on $K\times K$ and a maximiser at level $\alpha$ is a point at which it is attained, both provided by Limits of Penalised Maxima on a Compact Subset of a Metric Space §attainment, and $M$ is the greatest lower bound of the set of the $M_{\alpha}$, provided by Limits of Penalised Maxima on a Compact Subset of a Metric Space §infimum.

By Limits of Penalised Maxima on a Compact Subset of a Metric Space §infimum we have $\Phi(z)\le M$ for every $z\in Z$, that is

Since adding $v(x)$ to both sides turns $u(x)-v(x)\le 0$ into $u(x)\le v(x)$, it suffices to prove that $M\le 0$.

Step 2 (Assume $0<M$; choice of scales). Suppose, seeking a contradiction, that $M\le 0$ fails. Since $\le$ is a total order, $0<M$, so $\gamma M$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field and $\tfrac{\gamma M}{2}$ is positive by claim 8 of that lemma.

Apply Modulus of Continuity with $\varepsilon=\tfrac{\gamma M}{2}$: there is a positive $\eta_{0}\in\mathbb{R}$ such that every $t\in T$ with $t\le\eta_{0}$ satisfies $\omega(t)\le\tfrac{\gamma M}{2}$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\eta\in\mathbb{R}$ with $\eta\le\eta_{0}$, $\eta\le 1$ and $\eta$ equal to $\eta_{0}$ or to $1$; in either case $\eta$ is positive, and every $t\in T$ with $t\le\eta$ satisfies $t\le\eta_{0}$ and hence $\omega(t)\le\tfrac{\gamma M}{2}$.

Put $\theta=\tfrac{\eta}{2}$ and $\varepsilon^{\ast}=\tfrac{\theta^{2}}{2}$, where $\theta^{2}=\theta\theta$. By claim 8 of Elementary Order Arithmetic in an Ordered Field, $\theta$ is positive, $\theta<\eta$ and $\theta+\theta=\eta$; by claim 5 of that lemma $\theta^{2}$ is positive, and by claim 8 again $\varepsilon^{\ast}$ is positive with $\varepsilon^{\ast}+\varepsilon^{\ast}=\theta^{2}$. Moreover $\theta<\eta\le1$, so multiplying $\theta\le 1$ by the nonnegative number $\theta$ gives $\theta^{2}\le\theta$.

Apply Limits of Penalised Maxima on a Compact Subset of a Metric Space §penalty-vanishes with $\varepsilon=\varepsilon^{\ast}$: there is $\alpha_{0}\in\mathbb{R}$ with $1\le\alpha_{0}$ such that every $\alpha\in\mathbb{R}$ with $\alpha_{0}\le\alpha$ and every maximiser $(x,y)$ at level $\alpha$ satisfy

For such $\alpha$ and $(x,y)$, write $h=d_{E}(x,y)=\lVert x-y\rVert$, the second equality by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so that $2\Psi(x,y)=h^{2}$. Multiplying the first two inequalities by the positive number $2$ and using claim 10 of Elementary Order Arithmetic in an Ordered Field gives

hence $\alpha h^{2}<\theta$ by claim 2 of that lemma, and $h<\theta$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, both $h$ and $\theta$ being nonnegative. Adding these two strict inequalities with claim 3 of Elementary Order Arithmetic in an Ordered Field yields

and since $\alpha h^{2}+h$ lies in $T$, as observed in the statement, we conclude that $\omega\bigl(\alpha h^{2}+h\bigr)\le\tfrac{\gamma M}{2}$.

Step 3 (An interior maximiser). Let $\iota_{\mathbb{N}}:\mathbb{N}\to\mathbb{R}$ be the map of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and put $\alpha_{k}=\alpha_{0}+\iota_{\mathbb{N}}(k)$ for $k\in\mathbb{N}$. By claim 2 of that lemma $1\le\iota_{\mathbb{N}}(k)$, so each $\alpha_{k}$ is positive and $\alpha_{0}\le\alpha_{k}$. Given $R\in\mathbb{R}$, claim 2 of The Archimedean Property of the Real Numbers provides $N\in\mathbb{N}$ with $R-\alpha_{0}<\iota_{\mathbb{N}}(N)$; for $k\in\mathbb{N}$ with $N\le k$ we have $\iota_{\mathbb{N}}(N)\le\iota_{\mathbb{N}}(k)$, by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field if $N<k$ and trivially if $N=k$, so $R-\alpha_{0}<\iota_{\mathbb{N}}(k)$ and hence $R<\alpha_{k}$.

For each $k\in\mathbb{N}$ choose a maximiser $(x_{k},y_{k})$ at level $\alpha_{k}$, which exists by Limits of Penalised Maxima on a Compact Subset of a Metric Space §attainment. By Limits of Penalised Maxima on a Compact Subset of a Metric Space §cluster-points the sequence $\bigl((x_{k},y_{k})\bigr)_{k\in\mathbb{N}}$ has a cluster point $(\hat{x},\hat{y})$ lying in $K\times K$, and every such cluster point satisfies $\Psi(\hat{x},\hat{y})=0$ and $\Phi(\hat{x},\hat{y})=M$. By Step 1 the first gives $\hat{x}=\hat{y}$, and the second then reads $u(\hat{x})-v(\hat{x})=M$.

Since $0<M$, adding $v(\hat{x})$ gives $v(\hat{x})<u(\hat{x})$, so the hypothesis $u(x)\le v(x)$ for $x\in\partial\Omega$ shows $\hat{x}\notin\partial\Omega$. As $\overline{\Omega}=\Omega\cup\partial\Omega$ by Bounded Open Domain in Euclidean Space §boundary, we get $\hat{x}\in\Omega$. Since $\Omega$ is open, Open Subset of a Metric Space provides a positive $\rho\in\mathbb{R}$ such that every $x\in\mathbb{R}^{n}$ with $d_{E}(x,\hat{x})<\rho$ lies in $\Omega$.

By Cluster Point of a Sequence in a Metric Space, applied with $\varepsilon=\rho$, there is $k\in\mathbb{N}$ with $d_{\times}\bigl((x_{k},y_{k}),(\hat{x},\hat{x})\bigr)<\rho$; by claim 3 of The Product Metric is a Metric this gives $d_{E}(x_{k},\hat{x})<\rho$ and $d_{E}(y_{k},\hat{x})<\rho$, so $x_{k},y_{k}\in\Omega$. Fix such a $k$ and write $\alpha=\alpha_{k}$, $x=x_{k}$, $y=y_{k}$ and $h=\lVert x-y\rVert$; the conclusions of Step 2 apply to $\alpha$ and $(x,y)$ because $\alpha_{0}\le\alpha$.

Step 4 (Ishii's lemma at $(x,y)$). By Viscosity Subsolution and Supersolution up to the Boundary, $u|_{\Omega}$ is a viscosity subsolution and $v|_{\Omega}$ a viscosity supersolution of $F$ on $\Omega$; in particular, by Viscosity Subsolution and Supersolution of a Second-Order Equation, $u|_{\Omega}$ is upper semicontinuous on $\Omega$ and $v|_{\Omega}$ is lower semicontinuous on $\Omega$.

Since $(x,y)$ is a maximiser at level $\alpha$ and $\Omega\subseteq\overline{\Omega}$, every $x',y'\in\Omega$ satisfy

using claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. The hypothesis of Ishii's Lemma: Test Data and Matrix Bounds at a Maximum of a Quadratically Penalised Difference therefore holds for $\Omega$, the functions $u|_{\Omega}$ and $v|_{\Omega}$, the parameter $\alpha$, the points $x$ and $y$, and any positive $\delta$, for instance $\delta=1$. Let $X,Y\in\mathcal{S}(n)$ be as provided there and put $q=\alpha(x-y)$. Then Ishii's Lemma: Test Data and Matrix Bounds at a Maximum of a Quadratically Penalised Difference §test-data says that $\bigl(x,u(x),q,X\bigr)$ is approximable by test data from above for $u|_{\Omega}$ and that $\bigl(y,v(y),q,Y\bigr)$ is approximable by test data from below for $v|_{\Omega}$, and Ishii's Lemma: Test Data and Matrix Bounds at a Maximum of a Quadratically Penalised Difference §quadratic-bound says that

Since $F$ is continuous, it is in particular continuous at the two quadruples just named, so Viscosity Inequalities Pass to Limits of Test-Function Data §sub and Viscosity Inequalities Pass to Limits of Test-Function Data §super give

Step 5 (The contradiction). Write $r=u(x)$ and $s=v(y)$. By Step 2, $M\le\Phi(x,y)=r-s$, and $0<M$, so $0<r-s$ and hence $s\le r$.

By condition 2 of the statement and clause 2 of Strictly Proper Second-Order Equation Operator, applied at the point $x$ with the vector $q$, the matrix $X$ and the pair $s\le r$,

Now

The first summand is at most $0$, by Step 4. The second is at most $\omega\bigl(\alpha h^{2}+h\bigr)$ by condition 3 of the statement, applied with the points $x$ and $y$, the value $s$, the parameter $\alpha$ and the matrices $X$ and $Y$, whose hypothesis is the two-sided bound of Step 4 and in which $\alpha(x-y)=q$ and $\lVert x-y\rVert=h$. Adding the two bounds gives

Finally, multiplying $M\le r-s$ by the nonnegative number $\gamma$ gives $\gamma M\le\gamma(r-s)$, so combining the last three displays with Step 2,

Adding $-\tfrac{\gamma M}{2}$ to both sides and using $\tfrac{\gamma M}{2}+\tfrac{\gamma M}{2}=\gamma M$, from claim 8 of Elementary Order Arithmetic in an Ordered Field, gives $\tfrac{\gamma M}{2}\le 0$. Together with the positivity of $\tfrac{\gamma M}{2}$ established in Step 2 this yields $0<0$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, which is false.

Therefore $M\le 0$, and by Step 1 we conclude that $u(x)\le v(x)$ for every $x\in\overline{\Omega}$.